Question

0.500 g of a sample which contains copper is analyzed iodometrically. Copper (II) is reduced to copper (I) by the iodide. What is the percent copper in the sample if 20.00 cm3 of 0.1 M sodium thiosulphate solution is required for the titration of the liberated iodine?

Answer 1

Answer:

                      %Copper  =  25.42 %

Explanation:

                    For solving this problem one first write down the half reactions as;

Reduction of Cu²⁺ to Cu⁺¹ by I⁻:

                                2 Cu²⁺  +  4 I⁻   →   2 CuI  +  I₂

Secondly, the titration reaction between Na₂S₂O₃ with I₂ as;

                             2 S₂O₃²⁻  +  I₂   →   S₄O₆²⁻  +  2 I⁻

Now, Calculate Moles of S₂O₃²⁻ as;

                             Moles  =  Molarity × Volume

Putting values,

                             Moles  =  0.1 mol/L × 0.02 L

                             Moles  =  0.002 moles of S₂O₃²⁻

From first two equations we can conclude that,

2 moles of S₂O₃²⁻ liberated I₂ produced by  =  2 moles of Cu²⁺

So,

0.002 moles of S₂O₃²⁻ will liberate I₂ produced by  = X moles of Cu₂⁺

Solving for X,

                       X =  0.002 mol × 2 mol / 2 mol

                       X  =  0.002 mol of Cu²⁺

So, we require 0.002 moles of Cu²⁺ to produce I₂ enough for titrating with 0.1 M (20.00 mL) sodium thiosulphate solution.

Now, Calculate mass of Copper as,

                       Mass  =  Moles  ×  A.Mass

                       Mass  =  0.002 mol × 63.55 g/mol

                       Mass  =  0.1271 g of Copper

Lastly, calculate % copper as,

                       %Copper  =  0.1271 g / 0.500 g × 100

                       %Copper  =  25.42 %